Intuitively it feels like the answer to your question ought to be "yes", but my attempt to prove it has foundered. But it has suggested a stupid question. In Euclidean space of dimension, a sphere of radius r (meaning, the set of points at distance r from a given center) has curvature r^(1-d). What is the corresponding formula for spheres of radius r in spaces of constant non-zero curvature? Obviously it never goes negative, because that would make the "sphere" non-compact. On Thu, Apr 7, 2016 at 1:13 PM, Mike Stay <metaweta@gmail.com> wrote:
Yeah, sorry, mixed up the signs on the curvatures.
On Thu, Apr 7, 2016 at 10:05 AM, Allan Wechsler <acwacw@gmail.com> wrote:
The way you started to state the problem is a little confusing to me. You start by giving an example of a space being embedded in a higher-dimensional one with *lower* curvature. And then the hyperbolic surfaces ought to have been presented as having curvatures of -1 and -2, and again, the higher dimensional one has lower curvature. Perhaps you just meant to say "lower curvature and more dimensions" in the second sentence?
On Thu, Apr 7, 2016 at 12:55 PM, Mike Stay <metaweta@gmail.com> wrote:
A sphere can be embedded into Euclidean 3-space. Can a lower-curvature space always be embedded in a space with higher curvature and more dimensions? I.e. could a hyperbolic surface with constant curvature 1 be embedded in a three-dimensional hyperbolic space with constant curvature 2? -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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